Class 12th

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New answer posted

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

l = 1 2 c ε 0 E 0 2 ⇒ 8 4 π ( 1 0 ) 2 * 1 2 = 1 4 * c * 1 c 2 μ 0 * E 0 2 ⇒ E 0 = 2 1 0 μ 0 c π ⇒ x = 2

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

After contact, charges an each sphere will be

q1+q22=1nc⇒F=kq1q2r2=36*10−9N

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Energy corresponding to a particle is mc2

hcλ=mc2⇒hcλ= (x3h)c2⇒x=3λc=310*10−10*3*108 = 110−1=10

New answer posted

a year ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

i2 = 2i1

i1 + i2 = 6

i1 = 2A.

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

Tsinθ=kq2r2

Tcosθ=mg

tanθ=kq2mgr2

Using the above equation find the value of 'q'

New question posted

a year ago

0 Follower 4 Views

New answer posted

a year ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

Since force will act in horizontal direction so vertical component of speed will be same

v1 cos α = v2 cos β

⇒ K 1 K 2 = ( v 1 v 2 ) 2 = c o s 2 β c o s 2 α

New answer posted

a year ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

In ferromagnetic material, below Curie's temperature, a domain is defined as macroscopic region with zero magentisation.

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Using Bohr's theory :

λ=hcE2−E1=1242eV−nm (−3.4)− (−13.6)=121.8nm

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Using de-Broglie equation for wave nature of particle :

λ=hcmv⇒λα1m

λeλP=1me1mP=mPme=1836

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